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Question
A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = `100 sqrt2` sin (1000 t). The power factor of the combination is ______.
Options
`1/sqrt2`
`1/sqrt3`
0.5
0.6
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Solution
A resistor of 500 Ω and inductance of 0.5 H is in series with an AC source which is given by V = `100 sqrt2` sin (1000 t). The power factor of the combination is `1/sqrt2`.
Explanation:
Substitute the given values from
L = 0.5 H .....(1)
R = 500 Ω .....(2)
V = `100sqrt(2) sin(1000t)` .....(3)
On comparing with, V = `V_0 sin(wt)`
We get, `w = 1000s^-1` .....(4)
Put the values in equation (1) and (4) and calculate:
Let, Inductive reactance XL
XL = `w_L` = 1000 × 0.5
XL = 500 Ω ......(5)
Since, Power factor `cos phi`
`cos phi = R/sqrt(X_L^2 + R^2)` ....(7)
∴ `cos phi = 500/sqrt(500^2 + 500^2)`
`cos phi = 500/(500sqrt(2)`
`cos phi = 1/sqrt(2)`
Therefore, P.F = `cos phi = 1/sqrt(2)`
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